<h3>
Answer:</h3>
#a. OH⁻(aq) + H⁺(aq) → H₂O(l)
#b. 0.000988 moles
#c. 0.000988 moles
#d.30.875 L
<h3>
Solution and explanation:</h3>
38.0 mL of a 0.026 M of HCl is needed to react completely with 0.032 M NaOH solution.
<h3>#a. Net ionic equation for the reaction</h3>
The reaction between NaOH and HCl is known as neutralization reaction which results to the formation of salt and water as the product.
There equation for the reaction is;
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
The net ionic equation for a reaction involves only the ions that have changed their state.
The complete ionic equation is;
Na⁺(aq)OH⁻(aq) + H⁺(aq)Cl⁻(aq) → Na⁺(aq)Cl(aq) + H₂O(l)
Therefore, the net ionic equation will be;
OH⁻(aq) + H⁺(aq) → H₂O(l)
<h3>#b. Moles of HCl originally present.</h3>
Number of moles is given by multiplying the volume of the solution by its molarity or concentration.
Moles = Volume × Concentration
Volume of HCl = 38.0 mL
Concentration = 0.026 M
Therefore;
Moles = 0.038 × 0.026
= 0.000988 moles
Moles of HCl is 0.000988 moles
<h3>
#c. Moles of NaOH used in the reaction </h3>
From the equation , 1 mole of HCl reacts with 1 mole of NaOH
Therefore;
The mole ratio of HCl : NaOH is 1 : 1
Hence, moles of NaOH will be 0.000988 moles
0.000988 moles were neutralized by the acid.
<h3>
#d. Volume of NaOH required to react with the acid</h3>
Moles = Volume × Molarity
Therefore;
Volume = Moles ÷ Molarity
Moles of NaOH is 0.000988 moles
Molarity of NaOH is 0.032 M
Thus;
Volume of NaOH = 0.000988 moles ÷ 0.032 M
= 0.030875 mL
= 30.875 L
Therefore, the volume of the NaOH solution required to react with the acid is 30.875 L.
